Lemma

Proof

It is clear that the third condition implies the first two: take the section/retract to be given by the canonical injection/projection maps that come with a direct sum.

Conversely, suppose we have a retract r:B→Ar \colon B \to A of i:A→Bi \colon A \to B. Write P:B→rA→iBP \colon B \stackrel{r}{\to} A \stackrel{i}{\to} B for the corresponding idempotent.

Then every element b∈Bb \in B can be decomposed as b=(b−P(b))+P(b)b = (b - P(b)) + P(b) hence with b−P(b)∈ker(r)b - P(b) \in ker(r) and P(b)∈im(i)P(b) \in im(i). Moreover this decomposition is unique since if b=i(a)b = i(a) while at the same time r(b)=0r(b) = 0 then 0=r(i(a))=a0 = r(i(a)) = a. This shows that B≃im(i)⊕ker(r)B \simeq im(i) \oplus ker(r) is a direct sum and that i:A→Bi \colon A \to B is the canonical inclusion of im(i)im(i). By exactness it then follows that ker(r)≃im(p)ker(r) \simeq im(p) and hence that B≃A⊕CB \simeq A \oplus C with the canonical inclusion and projection.

The implication that the second condition also implies the third is formally dual to this argument.